1
WB JEE 2022
+2
-0.5

If the transformation $$z = \log \tan {x \over 2}$$ reduces the differential equation

$${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$ into the form $${{{d^2}y} \over {d{z^2}}} + ky = 0$$ then k is equal to

A
$$-$$4
B
4
C
2
D
$$-$$2
2
WB JEE 2021
+1
-0.25
The differential equation of all the ellipses centred at the origin and have axes as the co-ordinate axes is where $$y^{\prime}\equiv{{{dx}\over {dy}}},y^{\prime\prime}\equiv{{{d^2}y\over {dx^2}}}$$
A
y2 + xy'2 $$-$$ yy' = 0
B
xyy'' + xy'2 $$-$$ yy' = 0
C
yy' + xy'2 $$-$$ xy' = 0
D
x2y' + xy'' $$-$$ 3y = 0
3
WB JEE 2021
+1
-0.25
If $$x{{dy} \over {dx}} + y = {{xf(xy)} \over {f'(xy)'}}$$, then | f(xy) | is equal to (where k is an arbitrary positive constant).
A
$$k{e^{{x^2}/2}}$$
B
$$k{e^{{y^2}/2}}$$
C
$$k{e^{{x^2}}}$$
D
$$k{e^{{y^2}}}$$
4
WB JEE 2021
+2
-0.5
The differential of $$f(x) = {\log _e}(1 + {e^{10x}}) - {\tan ^{ - 1}}({e^{5x}})$$ at x = 0 and for dx = 0.2 is
A
0.5
B
0.3
C
$$-$$ 0.2
D
$$-$$ 0.5
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